Solution
Let say , $$x=\sqrt{6+\sqrt{6+\sqrt{6............\ }\ }}.$$
So, $$x^2=6+\sqrt{6+\sqrt{6............\ }}=6+x.$$
or,$$x^2-x-6=0.$$
or,$$x^2-3x+2x-6=0.$$
or,$$x\left(x-3\right)+2\left(x-3\right)=0.$$
or,$$\left(x+2\right)\left(x-3\right)=0.$$
So, either x=3 or x=-2.
But negative value is possible.
A is correct choice.
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